4 Answers
4

Substitute $u=1/x$ to get
$$
\int \frac{u^3 - u}{\sqrt{u^4-2u^2+2}}\,du
$$
This integral is much simpler, and can be solved by substituting $v = u^4 - 2u^2 + 2$. The final result is
$$
\int\frac{x^{2}-1}{x^{3}\sqrt{2x^{4}-2x^{2}+1}}\,dx \;=\; \frac{\sqrt{2x^4-2x^2+1}}{2x^2} + C.
$$

The factor of $(2x^4 - 2x^2 + 1)^{-1/2}$ suggests that it might be profitable to look at solutions of the form $f(x)(2x^4 - 2x^2 + 1)^{1/2}$, and hope for a simplification. Indeed, by differentiating this expression we get the differential equation